找到特定组中最常见的值

Question

我有一个我需要修改的查询.它目前每个参考都有一行,15分钟的时间段和相关的帐号.我的目标是每15分钟阻止一行,说明在该阻止期间最常出现的帐号.

    SELECT
left(cast(Concat([Hour],':',[Minute]) as time),5) as [Time Block]
,[Reference]
,[Account]

from(

    SELECT 
DATEPART(hh,[Start Time]) as [Hour]
,[Reference]
,case when DATEPART(mi,[Start Time]) between 00 and 15 then '00'
when DATEPART(mi,[Start Time]) between 15 and 30 then '15'
        when DATEPART(mi,[Start Time]) between 30 and 45 then '30'
        when DATEPART(mi,[Start Time]) between 45 and 100 then '00'
        else 'Error' end AS [Minute]

      ,[Account]

  FROM [iPR].[dbo].[InboundCallsView]

  where [Start Time] between '2017-03-21' and '2017-03-22')T

  order by [Time Block]

这给出了输出

我不需要参考号码,但我想要

00:00 310523

00:15 310523

00:30 310523

依此类推24小时内每15分钟显示最多行的帐户.这可能吗？

Answer 1

该值在统计中称为模式.它很容易计算:

with cte as (<your query here>)
select timeblock, account
from (select timeblock, account, count(*) as cnt,
             row_number() over (partition by timeblock order by count(*) desc) as seqnum
      from cte
      group by timeblock, account
     ) t
where seqnum = 1;

如果是最常见的联系,则会任意返回一个值.如果你想要所有这些,那么使用rank()或dense_rank().